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Question Number 137048 by mnjuly1970 last updated on 29/Mar/21
........advanced ... ... ... calculus....... evaluate:::: π›˜=∫_0 ^( ∞) x^2 e^(βˆ’x^2 ) ln(x)=???
$$\:\:\:\:\:\:\:\:\:\:\:……..{advanced}\:\:…\:…\:…\:{calculus}……. \\ $$$$\:\:\:\:{evaluate}:::: \\ $$$$\:\:\:\:\:\:\boldsymbol{\chi}=\int_{\mathrm{0}} ^{\:\infty} {x}^{\mathrm{2}} \:{e}^{βˆ’{x}^{\mathrm{2}} } {ln}\left({x}\right)=??? \\ $$$$\:\: \\ $$
Answered by Ar Brandon last updated on 29/Mar/21
π›˜=∫_0 ^∞ x^2 e^(βˆ’x^2 ) lnxdx=(βˆ‚/βˆ‚t)∣_(t=0) ∫_0 ^∞ x^(2+t) e^(βˆ’x^2 ) dx =^((u=x^2 )) (βˆ‚/βˆ‚t)∣_(t=0) (1/2)∫_0 ^∞ u^((1/2)+(t/2)) e^(βˆ’u) du=(βˆ‚/βˆ‚t)∣_(t=0) (1/2)Ξ“((3/2)+(t/2)) =(1/2)∣_(t=0) (1/2)Ξ“((3/2)+(t/2))ψ((3/2)+(t/2)) =(1/4)Ξ“((3/2))ψ((3/2))=(1/4)βˆ™((βˆšΟ€)/2)βˆ™(2βˆ’Ξ³βˆ’2ln2)
$$\boldsymbol{\chi}=\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{\mathrm{2}} \mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \mathrm{lnxdx}=\frac{\partial}{\partial\mathrm{t}}\mid_{\mathrm{t}=\mathrm{0}} \int_{\mathrm{0}} ^{\infty} \mathrm{x}^{\mathrm{2}+\mathrm{t}} \mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\overset{\left(\mathrm{u}=\mathrm{x}^{\mathrm{2}} \right)} {=}\frac{\partial}{\partial\mathrm{t}}\mid_{\mathrm{t}=\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \mathrm{u}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{t}}{\mathrm{2}}} \mathrm{e}^{βˆ’\mathrm{u}} \mathrm{du}=\frac{\partial}{\partial\mathrm{t}}\mid_{\mathrm{t}=\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{t}}{\mathrm{2}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mid_{\mathrm{t}=\mathrm{0}} \frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{t}}{\mathrm{2}}\right)\psi\left(\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{t}}{\mathrm{2}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\sqrt{\pi}}{\mathrm{2}}\centerdot\left(\mathrm{2}βˆ’\gammaβˆ’\mathrm{2ln2}\right) \\ $$
Commented by mnjuly1970 last updated on 29/Mar/21
very very nice mr brandon ..thank you...
$$\:{very}\:{very}\:{nice} \\ $$$$\:\:{mr}\:{brandon}\:..{thank}\:{you}… \\ $$
Commented by Ar Brandon last updated on 29/Mar/21
My pleasure, Sir
Answered by Dwaipayan Shikari last updated on 29/Mar/21
∫_0 ^∞ x^2 e^(βˆ’x^2 ) log(x)dx=β„΅β€²(3) β„΅(Ξ±)=∫_0 ^∞ x^(Ξ±βˆ’1) e^(βˆ’x^2 ) dxβ‡’β„΅(Ξ±)=(1/2)∫_0 ^∞ j^((Ξ±/2)βˆ’1) e^(βˆ’j) dj β„΅(Ξ±)=((Ξ“((Ξ±/2)))/2)β‡’β„΅β€²(Ξ±)=((Ξ“β€²((Ξ±/2)))/4)β‡’β„΅β€²(3)=((βˆšΟ€)/8)(βˆ’Ξ³+2βˆ’2log(2))
$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{βˆ’{x}^{\mathrm{2}} } {log}\left({x}\right){dx}=\aleph'\left(\mathrm{3}\right) \\ $$$$\aleph\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} {x}^{\alphaβˆ’\mathrm{1}} {e}^{βˆ’{x}^{\mathrm{2}} } {dx}\Rightarrow\aleph\left(\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {j}^{\frac{\alpha}{\mathrm{2}}βˆ’\mathrm{1}} {e}^{βˆ’{j}} {dj} \\ $$$$\aleph\left(\alpha\right)=\frac{\Gamma\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}}\Rightarrow\aleph'\left(\alpha\right)=\frac{\Gamma'\left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{4}}\Rightarrow\aleph'\left(\mathrm{3}\right)=\frac{\sqrt{\pi}}{\mathrm{8}}\left(βˆ’\gamma+\mathrm{2}βˆ’\mathrm{2}{log}\left(\mathrm{2}\right)\right) \\ $$
Commented by mnjuly1970 last updated on 29/Mar/21
thanks alot mr payan...
$${thanks}\:{alot}\:{mr}\:{payan}… \\ $$
Commented by Dwaipayan Shikari last updated on 29/Mar/21
Have a great day sir!
Answered by mathmax by abdo last updated on 30/Mar/21
Ξ¦=∫_0 ^∞ x^2 e^(βˆ’x^2 ) lnx dx let f(a) =∫_0 ^∞ x^a e^(βˆ’x^2 ) dx v(o) x^a e^(βˆ’x^2 ) ∼x^a ∫_0 ^ΞΎ (dx/x^(βˆ’a) ) cv β‡”βˆ’a<1 β‡’a>βˆ’1 we have f(a)=∫_0 ^∞ e^(alogx) e^(βˆ’x^2 ) dx β‡’f^β€² (a)=∫_0 ^∞ x^a e^(βˆ’x^2 ) logx dx and f^β€² (2)=∫_0 ^∞ x^2 e^(βˆ’x^2 ) logx dx changement x^2 =t givef(a)=∫_0 ^∞ t^(a/2) e^(βˆ’t) (dt/(2(√t))) =(1/2)∫_0 ^∞ t^((aβˆ’1)/2) e^(βˆ’t) dt =(1/2)∫_0 ^∞ t^(((aβˆ’1)/2)+1βˆ’1) e^(βˆ’t) dt =(1/2)∫_0 ^∞ t^(((a+1)/2)βˆ’1) e^(βˆ’t) dt =(1/2)Ξ“(((a+1)/2)) β‡’f^β€² (a)=(1/4)Ξ“^β€² (((a+1)/2)) β‡’ f^β€² (2) =(1/4)Ξ“^β€² ((3/2)) =Ξ¦
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{2}} \:\mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \mathrm{lnx}\:\mathrm{dx}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{a}} \:\mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\mathrm{v}\left(\mathrm{o}\right)\:\:\mathrm{x}^{\mathrm{a}} \:\mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \sim\mathrm{x}^{\mathrm{a}} \:\:\:\:\int_{\mathrm{0}} ^{\xi} \:\:\frac{\mathrm{dx}}{\mathrm{x}^{βˆ’\mathrm{a}} }\:\mathrm{cv}\:\Leftrightarrowβˆ’\mathrm{a}<\mathrm{1}\:\Rightarrow\mathrm{a}>βˆ’\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{\mathrm{alogx}} \:\mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \mathrm{dx}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{\mathrm{a}} \:\mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \mathrm{logx}\:\mathrm{dx} \\ $$$$\mathrm{and}\:\mathrm{f}^{'} \left(\mathrm{2}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{2}} \:\mathrm{e}^{βˆ’\mathrm{x}^{\mathrm{2}} } \mathrm{logx}\:\mathrm{dx} \\ $$$$\mathrm{changement}\:\mathrm{x}^{\mathrm{2}} =\mathrm{t}\:\mathrm{givef}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{a}}{\mathrm{2}}} \:\mathrm{e}^{βˆ’\mathrm{t}} \:\:\frac{\mathrm{dt}}{\mathrm{2}\sqrt{\mathrm{t}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{a}βˆ’\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{βˆ’\mathrm{t}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{a}βˆ’\mathrm{1}}{\mathrm{2}}+\mathrm{1}βˆ’\mathrm{1}} \:\mathrm{e}^{βˆ’\mathrm{t}} \mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}βˆ’\mathrm{1}} \:\mathrm{e}^{βˆ’\mathrm{t}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=\frac{\mathrm{1}}{\mathrm{4}}\Gamma^{'} \left(\frac{\mathrm{a}+\mathrm{1}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{2}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\Gamma^{'} \left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\Phi \\ $$$$ \\ $$

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